R中的三次样条插值

Question

我正在尝试在 R 中实现三次样条。我已经使用了 R 库中提供的 spline、smooth.spline 和 smooth.Pspline 函数，但我对结果不太满意，所以我想说服自己“自制”样条函数结果的一致性。我已经计算了三阶多项式的系数，但我不确定如何绘制结果......它们似乎是随机点。您可以在下面找到源代码。任何帮助将不胜感激。

x = c(35,36,39,42,45,48)
y = c(2.87671519825595, 4.04868309245341,   3.95202175000174,   
  3.87683188946186, 4.07739945984612,   2.16064840967985)


n = length(x)

#determine width of intervals
h=0
for (i in 1:(n-1)){
   h[i] = (x[i+1] - x[i])
}

A = 0
B = 0
C = 0
D = 0
#determine the matrix influence coefficients for the natural spline
for (i in 2:(n-1)){
  j = i-1
  D[j] = 2*(h[i-1] + h[i])
  A[j] = h[i]
  B[j] = h[i-1] 

}

#determine the constant matrix C
for (i in 2:(n-1)){
  j = i-1
  C[j] = 6*((y[i+1] - y[i]) / h[i] - (y[i] - y[i-1]) / h[i-1])
}

#maximum TDMA length
ntdma = n - 2

#tridiagonal matrix algorithm

#upper triangularization
R = 0
for (i in 2:ntdma){
  R = B[i]/D[i-1]
  D[i] = D[i] - R * A[i-1]
  C[i] = C[i] - R * C[i-1] 
}

#set the last C
C[ntdma] = C[ntdma] / D[ntdma]

#back substitute
for (i in (ntdma-1):1){
  C[i] = (C[i] - A[i] * C[i+1]) / D[i]
}

#end of tdma

#switch from C to S
S = 0
for (i in 2:(n-1)){
  j = i - 1
  S[i] = C[j]
}
#end conditions
S[1] <- 0 -> S[n]

#Calculate cubic ai,bi,ci and di from S and h
for (i in 1:(n-1)){
 A[i] = (S[i+ 1] - S[i]) / (6 * h[i])
 B[i] = S[i] / 2
 C[i] = (y[i+1] - y[i]) / h[i] - (2 * h[i] * S[i] + h[i] * S[i + 1]) / 6
 D[i] = y[i]
}


#control points
xx = c(x[2],x[4])
yy = 0
#spline evaluation
for (j in 1:length(xx)){
  for (i in 1:n){
    if (xx[j]<=x[i]){
      break
    }
    yy[i] = A[i]*(xx[j] - x[i])^3 + B[i]*(xx[j] - x[i])^2 + C[i]*(xx[j] - x[i]) + D[i]

 }
points(x,yy ,col="blue")
}

谢谢

Answer 1

好的，这里...

您的“控制点”是您要评估三次样条的点。因此返回的点数 (yy) 与 xx 的长度相同。这让我发现了一些事情：

for (j in 1:length(xx)){
  for (i in 1:n){
    if (xx[j]<=x[i]){
      break
    }
    yy[i] = A[i]*(xx[j] - x[i])^3 + B[i]*(xx[j] - x[i])^2 + C[i]*(xx[j] - x[i]) + D[i]

 }

这只是计算 yy 的“n”个值。你好，这是怎么了？它应该返回 length(xx) 值...

然后我想我发现了其他东西 - 你的“break”将退出 for 循环。你真正想要的是跳过 i 并继续下一个，直到你找到与你的观点相关的一个：

#spline evaluation
for (j in 1:length(xx)){
  for (i in 1:n){
    if (xx[j]<=x[i]){
      next
     }
    yy[j] = A[i]*(xx[j] - x[i])^3 + B[i]*(xx[j] - x[i])^2 + C[i]*(xx[j] - x[i]) + D[i]

 }
}

这是低效的，因为你正在计算一些 yy[j] 并在下一次循环中转储它们，但无论如何，它完成了工作。

将其包装到一个函数中，这样您就可以轻松地使用它。我的函数“myspline”采用 x 和 y 作为要拟合的数据，并采用 xx 向量作为预测位置。我可以这样做：

> xx=seq(35,48,len=100)
> yy = myspline(x,y,xx)
> plot(xx,yy,type="l")
> points(x,y)
> 

我得到一条穿过 (x,y) 点的漂亮曲线。除了第一点似乎错过了并趋向于零，所以我怀疑某处仍然存在差一错误。那好吧。 99% 完成。

这是代码：

myspline <- function(x,y,xx){

n = length(x)

h=0;yy=0
#determine width of intervals
for (i in 1:(n-1)){
   h[i] = (x[i+1] - x[i])
}

A = 0
B = 0
C = 0
D = 0
#determine the matrix influence coefficients for the natural spline
for (i in 2:(n-1)){
  j = i-1
  D[j] = 2*(h[i-1] + h[i])
  A[j] = h[i]
  B[j] = h[i-1] 

}

#determine the constant matrix C
for (i in 2:(n-1)){
  j = i-1
  C[j] = 6*((y[i+1] - y[i]) / h[i] - (y[i] - y[i-1]) / h[i-1])
}

#maximum TDMA length
ntdma = n - 2

#tridiagonal matrix algorithm

#upper triangularization
R = 0
for (i in 2:ntdma){
  R = B[i]/D[i-1]
  D[i] = D[i] - R * A[i-1]
  C[i] = C[i] - R * C[i-1] 
}

#set the last C
C[ntdma] = C[ntdma] / D[ntdma]

#back substitute
for (i in (ntdma-1):1){
  C[i] = (C[i] - A[i] * C[i+1]) / D[i]
}

#end of tdma

#switch from C to S
S = 0
for (i in 2:(n-1)){
  j = i - 1
  S[i] = C[j]
}
#end conditions
S[1] <- 0 -> S[n]

#Calculate cubic ai,bi,ci and di from S and h
for (i in 1:(n-1)){
 A[i] = (S[i+ 1] - S[i]) / (6 * h[i])
 B[i] = S[i] / 2
 C[i] = (y[i+1] - y[i]) / h[i] - (2 * h[i] * S[i] + h[i] * S[i + 1]) / 6
 D[i] = y[i]
}


#control points
#xx = seq(x[2],x[4],len=100)

#spline evaluation
for (j in 1:length(xx)){
  for (i in 1:n){
    if (xx[j]<=x[i]){
      next
     }
    yy[j] = A[i]*(xx[j] - x[i])^3 + B[i]*(xx[j] - x[i])^2 + C[i]*(xx[j] - x[i]) + D[i]
 }
}
return(yy)
}